A block of mass 5 kg is placed on a rough ho | vedclass

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(A) Given: Mass $m = 5 \ kg$,coefficient of friction $\mu = 0.5$,applied force $F = 60 \ N$,and $g = 10 \ ms^{-2}$.First,calculate the limiting fricti

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$7$

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(A) Given: Mass $m = 5 \ kg$,coefficient of friction $\mu = 0.5$,applied force $F = 60 \ N$,and $g = 10 \ ms^{-2}$.First,calculate the limiting frictional force $f_l$ acting on the block:$f_l = \mu N = \mu mg = 0.5 	imes 5 	imes 10 = 25 \ N$.Since the applied force $F = 60 \ N$ is greater than the limiting frictional force $f_l = 25 \ N$,the block will move.The net force acting on the block is $F_{net} = F - f_l = 60 - 25 = 35 \ N$.Using Newton's second law,$F_{net} = ma$,the acceleration $a$ is:$a = \frac{F_{net}}{m} = \frac{35}{5} = 7 \ ms^{-2}$.

$A$ block of mass $5 \ kg$ is placed on a rough horizontal surface having a coefficient of friction $0.5$. If a horizontal force of $60 \ N$ is acting on it,then the acceleration of the block is (Acceleration due to gravity,$g = 10 \ ms^{-2}$). (in $ms^{-2}$)

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